Lesson 18: Similarity and the Angle Bisector Theorem

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M2 GEOMETRY

Student Outcomes . Students state, understand, and prove the angle bisector theorem. . Students use the angle bisector theorem to solve problems.

Classwork Opening Exercise (5 minutes) The Opening Exercise should activate students’ prior knowledge acquired in Module 1 that is helpful in proving the angle bisector theorem. Scaffolding: Opening Exercise . If necessary, provide a. What is an angle bisector? visuals to accompany The bisector of an angle is a ray in the interior of the angle such that the two adjacent these questions. angles formed by it have equal measures. . For part (a):

b. Describe the angle relationships formed when parallel lines are cut by a transversal.

When parallel lines are cut by a transversal, the corresponding, alternate interior, and alternate exterior angles are all congruent; the same-side interior angles are supplementary.

c. What are the properties of an isosceles triangle? . For part (b): An isosceles triangle has at least two congruent sides, and the angles opposite to the congruent sides (i.e., the base angles) are also congruent.

Discussion (20 minutes) Prior to proving the angle bisector theorem, students observe the length relationships of the sides of a triangle when one of the angles of the triangle has been bisected. . For part (c): . In this lesson, we investigate the length relationships of the sides of a triangle when one angle of the triangle has been bisected. Provide students time to look for relationships between the side lengths. This requires trial and error on the part of students and may take several minutes.

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NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 18 M2 GEOMETRY

Discussion

In the diagram below, the angle bisector of ∠푨 in △ 푨푩푪 meets side 푩푪̅̅̅̅ at point 푫. Does the angle bisector create any observable relationships with respect to the side lengths of the triangle?

MP.7

Acknowledge any relationships students may find, but highlight the relationship 푩푫: 푪푫 = 푩푨: 푪푨. Then, continue the discussion below that proves this relationship.

. The following theorem generalizes our observation:

THEOREM: The angle bisector theorem: In △ 퐴퐵퐶, if the angle bisector of ∠퐴 meets side 퐵퐶̅̅̅̅ at point 퐷, then 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐴.

In words, the bisector of an angle of a triangle splits the opposite side into segments that have the same ratio as the adjacent sides.

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. Our goal now is to prove this relationship for all triangles. We begin by constructing a line through vertex 퐶 that is parallel to side ̅퐴퐵̅̅̅. Let 퐸 be the point where this parallel line meets the angle bisector, as shown.

Scaffolding: In place of the formal proof, students may construct angle bisectors for a series of triangles and take measurements to verify the relationship inductively.

If we can show that △ 퐴퐵퐷 ~ △ 퐸퐶퐷, then we can use what we know about similar triangles to prove the relationship 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐴. Consider asking students why it is that if they can show △ 퐴퐵퐷 ~ △ 퐸퐶퐷, they are closer to the goal of showing 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐴. Students should respond that similar triangles have proportional length relationships. The triangles 퐴퐵퐷 and 퐸퐶퐷, if shown similar, would give 퐵퐷: 퐵퐴 = 퐶퐷: 퐶퐸, three out of the four lengths needed in the ratio 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐴. . How can we show that △ 퐴퐵퐷 ~ △ 퐸퐶퐷? Provide students time to discuss how to show that the triangles are similar. Elicit student responses; then, continue with the discussion below. . It is true that △ 퐴퐵퐷 ~ △ 퐸퐶퐷 by the AA criterion for similarity. Vertical angles 퐴퐷퐵 and 퐸퐷퐶 are congruent and, therefore, equal. ∠퐷퐸퐶 and ∠퐵퐴퐷 are congruent and equal because they are alternate interior angles of parallel lines cut by a transversal. (Show the diagram below.)

. Since the triangles are similar, we know that 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐸. This is very close to what we are trying to show: 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐴. What must we do now to prove the theorem?  We have to show that 퐶퐸 = 퐶퐴. Once students have identified what needs to be done, that is, show that 퐶퐸 = 퐶퐴, provide them time to discuss how to show it. The prompts below can be used to guide students’ thinking.

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. We need to show that 퐶퐸 = 퐶퐴. Notice that the segments 퐶퐸 and 퐶퐴 are two sides of △ 퐴퐶퐸. How might that be useful?  If we could show that △ 퐴퐶퐸 is an isosceles triangle, then we would know that 퐶퐸 = 퐶퐴. . How can we show that △ 퐴퐶퐸 is an isosceles triangle?  We were first given that angle 퐴 was bisected by 퐴퐷̅̅̅̅, which means that ∠퐵퐴퐷 ≅ ∠퐶퐴퐷. Then, by alternate interior ∠’s, 퐶퐸̅̅̅̅ ∥ ̅퐴퐵̅̅̅, it follows that ∠퐶퐴퐷 = ∠퐶퐸퐴. We can use the converse of the base angles of isosceles triangle theorem (i.e., base ∠’s converse). Since ∠퐶퐴퐸 = ∠퐶퐸퐴, then △ 퐴퐶퐸 must be an isosceles triangle. . Now that we know △ 퐴퐶퐸 is isosceles, we can conclude that 퐶퐸 = 퐶퐴 and finish the proof of the angle bisector theorem. All we must do now is substitute 퐶퐴 for 퐶퐸 in 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐸. Therefore, 퐵퐷: 퐶퐷 = 퐵퐴: 퐶퐴, and the theorem is proved. Consider asking students to restate what was just proved and summarize the steps of the proof. Students should respond that the bisector of an angle of a triangle splits the opposite side into segments that have the same ratio as the adjacent sides.

Exercises 1–4 (10 minutes) Students complete Exercises 1–4 independently.

Exercises 1–4

1. The sides of a triangle are ퟖ, ퟏퟐ, and ퟏퟓ. An angle bisector meets the side of length ퟏퟓ. Find the lengths 풙 and 풚. Explain how you arrived at your answers.

풚 ퟏퟐ 풙 = ퟏퟓ − ퟗ = 풙 ퟖ 풙 = ퟔ 풙 = ퟏퟓ − 풚

ퟖ풚 = ퟏퟐ풙 ퟖ풚 = ퟏퟐ(ퟏퟓ − 풚) ퟖ풚 = ퟏퟖퟎ − ퟏퟐ풚 ퟐퟎ풚 = ퟏퟖퟎ 풚 = ퟗ

The length 풙 is ퟔ, and the length 풚 is ퟗ.

Since I know that ∠푨 is bisected, I applied what I knew about the angle bisector theorem to determine the lengths 풙 and 풚. Specifically, the angle bisector cuts the side that is opposite the bisected angle so that 풚: 풙 = ퟏퟐ: ퟖ. I set up an equation using the values of the ratios, which could be solved once I rewrote one of the variables 풙 or 풚. I rewrote 풙 as ퟏퟓ − 풚 and then solved for 풚. Once I had a value for 풚, I could replace it in the equation 풙 = ퟏퟓ − 풚 to determine the value of 풙.

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2. The sides of a triangle are ퟖ, ퟏퟐ, and ퟏퟓ. An angle bisector meets the side of length ퟏퟐ. Find the lengths 풙 and 풚.

풚 = ퟏퟐ − 풙 ퟒ 풚 = ퟏퟐ − ퟒ ퟐퟑ 풙 ퟖ ퟏퟗ = 풚 = ퟕ ퟏퟐ − 풙 ퟏퟓ ퟐퟑ ퟏퟓ풙 = ퟖ(ퟏퟐ − 풙) ퟏퟓ풙 = ퟗퟔ − ퟖ풙 ퟐퟑ풙 = ퟗퟔ ퟗퟔ ퟒ 풙 = = ퟒ ퟐퟑ ퟐퟑ ퟒ ퟏퟗ 풙 ퟒ 풚 ퟕ The length of is ퟐퟑ, and the length of is ퟐퟑ.

3. The sides of a triangle are ퟖ, ퟏퟐ, and ퟏퟓ. An angle bisector meets the side of length ퟖ. Find the lengths 풙 and 풚.

풚 ퟏퟓ ퟓ = 풚 = ퟖ − ퟑ 풙 ퟏퟐ ퟗ 풚 = ퟖ − 풙 ퟒ 풚 = ퟒ ퟗ ퟏퟓ풙 = ퟏퟐ풚 ퟏퟓ풙 = ퟏퟐ(ퟖ − 풙) ퟏퟓ풙 = ퟗퟔ − ퟏퟐ풙 ퟐퟕ풙 = ퟗퟔ ퟗퟔ ퟏퟓ ퟓ 풙 = = ퟑ = ퟑ ퟐퟕ ퟐퟕ ퟗ ퟓ ퟒ 풙 ퟑ 풚 ퟒ The length of is ퟗ , and the length of is ퟗ.

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4. The angle bisector of an angle splits the opposite side of a triangle into lengths ퟓ and ퟔ. The perimeter of the triangle is ퟑퟑ. Find the lengths of the other two sides.

Let 풛 be the scale factor of a similarity. By the angle bisector theorem, the side of the triangle adjacent to the segment of length ퟓ has length of ퟓ풛, and the side of the triangle adjacent to the segment of length ퟔ has length of ퟔ풛. The sum of the sides is equal to the perimeter.

ퟓ + ퟔ + ퟓ풛 + ퟔ풛 = ퟑퟑ ퟏퟏ + ퟏퟏ풛 = ퟑퟑ ퟏퟏ풛 = ퟐퟐ 풛 = ퟐ

ퟓ(ퟐ) = ퟏퟎ, and ퟔ(ퟐ) = ퟏퟐ The lengths of the other two sides are ퟏퟎ and ퟏퟐ.

Closing (5 minutes) . Explain the angle bisector theorem in your own words. . Explain how knowing that one of the angles of a triangle has been bisected allows you to determine unknown side lengths of a triangle.

Exit Ticket (5 minutes)

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Name Date

Lesson 18: Similarity and the Angle Bisector Theorem

Exit Ticket

1. The sides of a triangle have lengths of 12, 16, and 21. An angle bisector meets the side of length 21. Find the lengths 푥 and 푦.

1 1 2. The perimeter of △ 푈푉푊 is 22 . 푊푍⃗⃗⃗⃗⃗⃗ bisects ∠푈푊푉, 푈푍 = 2, and 푉푍 = 2 . Find 푈푊 and 푉푊. 2 2

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Exit Ticket Sample Solutions

1. The sides of a triangle have lengths of ퟏퟐ, ퟏퟔ, and ퟐퟏ. An angle bisector meets the side of length ퟐퟏ. Find the lengths 풙 and 풚. 풚 ퟏퟔ By the angle bisector theorem, = , and 풚 = ퟐퟏ − 풙, so 풙 ퟏퟐ

ퟐퟏ − 풙 ퟏퟔ = 풚 = ퟐퟏ − 풙 풙 ퟏퟐ 풚 = ퟐퟏ − ퟗ ퟏퟐ(ퟐퟏ − 풙) = ퟏퟔ풙 풚 = ퟏퟐ ퟐퟓퟐ − ퟏퟐ풙 = ퟏퟔ풙

ퟐퟓퟐ = ퟐퟖ풙 ퟗ = 풙

ퟏ ퟏ △ 푼푽푾 ퟐퟐ 푾풁⃗⃗⃗⃗⃗⃗⃗ ∠푼푾푽 푼풁 = ퟐ 푽풁 = ퟐ 푼푾 푽푾 2. The perimeter of is ퟐ. bisects , , and ퟐ. Find and . ퟐ 푼푾 By the angle bisector theorem, = , so 푼푾 = ퟐ풙 and 푽푾 = ퟐ. ퟓ풙 for ퟐ.ퟓ 푽푾 ퟏ 풙 ퟐퟐ some positive number . The perimeter of the triangle is ퟐ, so ퟐ + ퟐ. ퟓ + ퟐ풙 + ퟐ. ퟓ풙 = ퟐퟐ. ퟓ ퟒ. ퟓ + ퟒ. ퟓ풙 = ퟐퟐ. ퟓ ퟒ. ퟓ풙 = ퟏퟖ 풙 = ퟒ

푼푾 = ퟐ풙 = ퟐ(ퟒ) = ퟖ 푽푾 = ퟐ. ퟓ풙 = ퟐ. ퟓ(ퟒ) = ퟏퟎ

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Problem Set Sample Solutions

ퟏ ퟏ ퟓ ퟖ, ퟔ ퟔ 1. The sides of a triangle have lengths of , and ퟐ. An angle bisector meets the side of length ퟐ. Find the lengths 풙 and 풚. 풙 ퟓ ퟏ Using the angle bisector theorem, = , and 풚 = ퟔ − 풙, so 풚 ퟖ ퟐ 풙 ퟓ ퟏ = 풚 = ퟔ − 풙 ퟏ ퟔ − 풙 ퟖ ퟐ ퟐ ퟏ ퟏ ퟏ 풚 = ퟔ − ퟐ ퟓ (ퟔ − 풙) = ퟖ풙 ퟐ ퟐ ퟐ 풚 = ퟒ ퟏ ퟑퟐ − ퟓ풙 = ퟖ풙 ퟐ ퟏ ퟑퟐ = ퟏퟑ풙 ퟐ ퟏ 풙 = ퟐ ퟐ

ퟏ ퟏ ퟏퟎ ퟏퟔ ퟗ. ퟗ 풙 풚 2. The sides of a triangle are ퟐ, ퟐ, and An angle bisector meets the side of length . Find the lengths and .

ퟏ ퟏퟔ 풙 ퟐ By the angle bisector theorem, = ퟏ, and 풚 = ퟗ − 풙, so 풚 ퟏퟎ ퟐ 풙 ퟏퟔ. ퟓ 풚 = ퟗ − 풙 = ퟗ − 풙 ퟏퟎ. ퟓ 풚 = ퟗ − ퟓ. ퟓ ퟏퟎ. ퟓ풙 = ퟏퟔ. ퟓ(ퟗ − 풙) 풚 = ퟑ. ퟓ ퟏퟎ. ퟓ풙 = ퟏퟒퟖ. ퟓ − ퟏퟔ. ퟓ풙

ퟐퟕ풙 = ퟏퟒퟖ. ퟓ 풙 = ퟓ. ퟓ

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ퟏ 푫푬푭 푫푮̅̅̅̅ 푫푬 = ퟖ 푫푭 = ퟔ 푬푭 = ퟖ 푭푮 푬푮 3. In the diagram of triangle below, is an angle bisector, , , and ퟔ. Find and .

ퟏ 푫푮̅̅̅̅ 푫 푭푮: 푮푬 = 푭푫: 푬푫 푬푭 = ퟖ Since is the angle bisector of angle , by the angle bisector theorem. If ퟔ, then ퟏ 푭푮 = ퟖ − 푮푬 ퟔ . ퟏ (ퟖ − 푮푬) 푭푬 = 푮푬 + 푭푮 ퟔ ퟔ = ퟏ ퟐ 푮푬 ퟖ ퟖ = ퟒ + 푭푮 ퟏ ퟔ ퟔ ퟑ ퟖ − 푮푬 = ∙ 푮푬 ퟏ ퟔ ퟖ 푭푮 = ퟑ ퟏ ퟏퟒ ퟐ ퟖ = ∙ 푮푬 ퟔ ퟖ ퟒퟗ ퟖ ∙ = 푮푬 ퟔ ퟏퟒ ퟐퟖ = 푮푬 ퟔ ퟐ 푮푬 = ퟒ ퟑ

4. Given the diagram below and ∠푩푨푫 ≅ ∠푫푨푪, show that 푩푫: 푩푨 = 푪푫: 푪푨.

Using the given information, 푨푫̅̅̅̅ is the angle bisector of angle 푨. By the angle bisector theorem, 푩푫: 푪푫 = 푩푨: 푪푨, so 푩푫 푩푨 = 푪푫 푪푨 푩푫 ∙ 푪푨 = 푪푫 ∙ 푩푨 푩푫 푪푫 = 푩푨 푪푨

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5. The perimeter of triangle 푳푴푵 is ퟑퟐ 퐜퐦. 푵푿̅̅̅̅̅ is the angle bisector of angle 푵, 푳푿 = ퟑ 퐜퐦, and 푿푴 = ퟓ 퐜퐦. Find 푳푵 and 푴푵.

Since 푵푿̅̅̅̅̅ is an angle bisector of angle 푵, by the angle bisector theorem, 푿푳: 푿푴 = 푳푵: 푴푵; thus, 푳푵: 푴푵 = ퟑ: ퟓ. Therefore, 푳푵 = ퟑ풙 and 푴푵 = ퟓ풙 for some positive number 풙. The perimeter of the triangle is ퟑퟐ 퐜퐦, so

푿푳 + 푿푴 + 푴푵 + 푳푵 = ퟑퟐ ퟑ + ퟓ + ퟑ풙 + ퟓ풙 = ퟑퟐ ퟖ + ퟖ풙 = ퟑퟐ ퟖ풙 = ퟐퟒ 풙 = ퟑ ퟑ풙 = ퟑ(ퟑ) = ퟗ and ퟓ풙 = ퟓ(ퟑ) = ퟏퟓ

푳푵 = ퟗ 퐜퐦 and 푴푵 = ퟏퟓ 퐜퐦

6. Given 푪푫 = ퟑ, 푫푩 = ퟒ, 푩푭 = ퟒ, 푭푬 = ퟓ, 푨푩 = ퟔ, and ∠푪푨푫 ≅ ∠푫푨푩 ≅ ∠푩푨푭 ≅ ∠푭푨푬, find the perimeter of quadrilateral 푨푬푩푪.

푨푫̅̅̅̅ is the angle bisector of angle 푪푨푩, so by the angle bisector theorem, 푪푫: 푩푫 = 푪푨: 푩푨. ퟑ 푪푨 = ퟒ ퟔ ퟒ ∙ 푪푨 = ퟏퟖ 푪푨 = ퟒ. ퟓ

푨푭̅̅̅̅ is the angle bisector of angle 푩푨푬, so by the angle bisector theorem, 푩푭: 푬푭 = 푩푨: 푬푨. ퟒ ퟔ = ퟓ 푬푨 ퟒ ∙ 푬푨 = ퟑퟎ 푬푨 = ퟕ. ퟓ

The perimeter of quadrilateral 푨푬푩푪: 푨푬푩푪 = 푪푫 + 푫푩 + 푩푭 + 푭푬 + 푬푨 + 푨푪 푨푬푩푪 = ퟑ + ퟒ + ퟒ + ퟓ + ퟕ. ퟓ + ퟒ. ퟓ 푨푬푩푪 = ퟐퟖ

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7. If 푨푬̅̅̅̅ meets 푩푪̅̅̅̅ at 푫 such that 푪푫: 푩푫 = 푪푨: 푩푨, show that ∠푪푨푫 ≅ ∠푩푨푫. Explain how this proof relates to the angle bisector theorem.

This is a proof of the converse to the angle bisector theorem.

∠푪푬푨 ≅ ∠푩푨푫 Alt. Int. ∠’s 푪푬̅̅̅̅ ∥ 푨푩̅̅̅̅

∠푨푫푩 ≅ ∠푬푫푪 Vert. ∠’s

∆푪푫푬 ~∆푩푫푨 AA ~ 푪푫 푪푬 = Corr. sides of ~ △’s 푫푩 푩푨 푪푫 푪푨 Given = implies then that 푩푫 푩푨 푪푬 = 푪푨, so △ 푨푬푪 is isosceles.

∠푪푨푫 ≅ ∠푪푬푨 Base ∠’s

∠푪푨푫 ≅ ∠푩푨푫 Substitution

8. In the diagram below, 푬푫̅̅̅̅ ≅ 푫푩̅̅̅̅̅, 푩푬̅̅̅̅ bisects ∠푨푩푪, 푨푫 = ퟒ, and 푫푪 = ퟖ. Prove that △ 푨푫푩 ~ △ 푪푬푩.

Using given information, 푬푫 = 푫푩, so it follows that 푬푩 = ퟐ푫푩. Since 푩푬̅̅̅̅ bisects ∠푨푩푪, ∠푨푩푫 ≅ ∠푪푩푫. Also, 푪푫 푩푪 푩푪 ퟖ ퟐ by the angle bisector theorem, = , which means = = . 푨푫 푩푨 푩푨 ퟒ ퟏ 푬푩 푪푩 ퟐ Since = = , and ∠푨푩푫 ≅ ∠푪푩푫, △ 푨푫푩 ~ △ 푪푬푩 by the SAS criterion for showing similar triangles. 푫푩 푨푩 ퟏ

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